Postby TGTS0907129» Electric field intensity equals potential change

Prove that electric field intensity is equal to the rate of change of potential

1-

The work ?w done by the field is given by
                             
?w = -F?s

If we substitute in this equation the value of F from the defining equation for field intensity (F = Eq), we obtain
                             
?w = -qE?s

Hence, E = -?V/?s

Equations state an important relationship: The electrical field intensity at a point in an electricity field is equal to the negative of the potential gradient of the field at this point.

If the test charge  +q is moved in some direction that is not along the line of force, the work done is given by ?w = -qE cos = θ ?s, where θ is the angle between ?s  and the tangent to the potential difference between the ends of ?s is
                            
?V = ?w/q = -(E cosθ)?s

And the rate of charge of potential with distance in this direction is ?V/?s = -Ecosθ. Hence the component of the electric field intensity in any given direction is equal to the negative of the rate of change of potential with distance in that direction,
                     
E cosθ = -?V/?s
 

By TGTT23071246 on 9/28/2015 11:45:19 PM
TGTS0907129 on 9/28/2015 11:38:30 PM
Posts:527
Top

Return to Forum